See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

Step 1: Identify the structure. The molecule shown is: N≡C1—C2(=O1)(—CH2—C3H3). This is a compound with a nitrile group (N≡C1), a carbonyl group on C2 (C2=O1), a methylene group (CH2), and a methyl-like carbon C3 bonded to three hydrogens. Step 2: Determine hybridization based on bonding and geometry. (i) N: Nitrogen is part of a triple bond (N≡C1). Atoms involved in triple bonds use sp hybridization. N has one triple bond and one lone pair in sp orbital → sp hybridized. (ii) C1: Carbon 1 is involved in a triple bond with N (N≡C1) and a single bond with C2. Two regions of electron density (triple bond counts as one) → sp hybridized. (iii) C2: Carbon 2 is bonded to C1 (single bond), O1 (double bond), and CH2 (single bond). Three regions of electron density → sp2 hybridized. (iv) O1: Oxygen is double-bonded to C2 (carbonyl oxygen). A carbonyl oxygen has two lone pairs and one double bond; the double bond involves one sigma and one pi bond. The oxygen has 3 regions of electron density (1 double bond + 2 lone pairs) but in practice carbonyl oxygen is considered sp2 hybridized because it participates in conjugation/resonance with the pi system of the C=O bond. (v) CH2: The methylene carbon is bonded to C2, C3, and two H atoms. Four single bonds → four regions of electron density → sp3 hybridized. (vi) C3: Carbon 3 is bonded to CH2, and three H atoms (shown as H, H, H on C3). Four single bonds → sp3 hybridized. Step 3: Why other options fail — any assignment other than sp for atoms in triple bonds, sp2 for carbonyl carbon/oxygen, and sp3 for saturated carbons would not match the observed geometry and bond angles. Therefore, the correct answer is {"i": "sp", "ii": "sp", "iii": "sp2", "iv": "sp2", "v": "sp3", "vi": "sp3"}.